Question

我正在尝试同时更新2个表tbl_loanledger和tbl_journal。两个表都具有相同的列，例如modified_by, date_modified and deleted。

当我尝试运行此操作时，我在此输入代码我收到错误：

错误号码：1052

字段列表中的'modified_by'列不明确

更新`loan_ledger`作为a，日期为b SET`modified_by` ='1'，   `date_modified` ='2016-06-08'，`删除`= 1 WHERE`a``id` ='823'   AND`b``id` ='823'

文件名：C：\ xampp \ htdocs \ system \ system \ database \ DB_driver.php

行号：331

这是如何在codeigniter中更新2个表的正确方法吗？

模型

public function delete($id){

  $data = array(
        'modified_by' => $this->ion_auth->user()->row()->id,
        'date_modified' => date("Y-m-d"),
        'deleted' => 1
      );

    $this->db->set($data);
    $this->db->where('a.id', $id);
    $this->db->where('b.id', $id);
    $this->db->update('loan_ledger as a, journal as b');

}

但是，如果我只是更新单个表，它正在运行，这是一个代码：

public function delete($id){

  $data = array(
        'modified_by' => $this->ion_auth->user()->row()->id,
        'date_modified' => date("Y-m-d"),
        'deleted' => 1
      );
    $this->db->where('id', $id);
    $this->db->update(tbl_loanledger, $data);

}

Answer 1

否使用连接只需将两个单独的更新查询写为

 // first 
    $this->db->set('a.modified_by', $this->ion_auth->user()->row()->id);
    $this->db->set('a.date_modified', date("Y-m-d"));
    $this->db->set('a.deleted', 1);
    $this->db->where('a.id',  $id);
    $this->db->update('loan_ledger as a');
    // second
    $this->db->set('b.modified_by', $this->ion_auth->user()->row()->id);
    $this->db->set('b.date_modified', date("Y-m-d"));
    $this->db->set('b.deleted', 1);
    $this->db->where('b.id',  $id);
    $this->db->update('journal as b');

Answer 2

试试这个

public function delete($id){

$data = array(
    'a.modified_by' => $this->ion_auth->user()->row()->id,
    'a.date_modified' => date("Y-m-d"),
    'b.modified_by' => $this->ion_auth->user()->row()->id,
    'b.date_modified' => date("Y-m-d"),  
    'deleted' => 1
  );

$this->db->set($data);
$this->db->where('a.id', $id);
$this->db->where('b.id', $id);
$this->db->update('loan_ledger as a, journal as b');

}

Answer 3

<?php
$this->db->set('a.modified_by', $this->ion_auth->user()->row()->id);
$this->db->set('a.date_modified', date("Y-m-d"));
$this->db->set('a.deleted', 1);

$this->db->set('b.modified_by', $this->ion_auth->user()->row()->id);
$this->db->set('b.date_modified', date("Y-m-d"));
$this->db->set('b.deleted', 1);

$this->db->where('a.id',  $id);
$this->db->where('a.id = b.id');
$this->db->update('loan_ledger as a, journal as b');
?>

更新两个表Codeigniter

3 个答案: